241.0 KB - NZTA Education Portal

... 1. Change the slope of the plank. Use stacking blocks or books to change the starting height of the trolley car. Make a prediction about any change in the trolley car’s motion before testing. Repeat your experiment three times, taking the average result under each condition as the speed or distance ...

... 1. Change the slope of the plank. Use stacking blocks or books to change the starting height of the trolley car. Make a prediction about any change in the trolley car’s motion before testing. Repeat your experiment three times, taking the average result under each condition as the speed or distance ...

The Study of Human Movement by the Application of Biomechanical

... weight of a barbell, that would cause the fixed lever arm to rotate about the joint. The weight (Force R) could also be supplied by an internal mass, as in the weight of an arm acting on the fixed lever arm (bone), causing it to rotate about the joint. Force E is the amount of weight a muscle produc ...

... weight of a barbell, that would cause the fixed lever arm to rotate about the joint. The weight (Force R) could also be supplied by an internal mass, as in the weight of an arm acting on the fixed lever arm (bone), causing it to rotate about the joint. Force E is the amount of weight a muscle produc ...

Holt Physics Problem 4D

... Choose the equation(s) or situation: Use Newton’s second law to describe the forces acting on the bicycle. Fnet = m anet = −Fk Use the definition of frictional force to express Fk in terms of the coefficient of friction. Fk = mk Fn = mk (mg) Rearrange the equation(s) to isolate the unknown(s): m ane ...

... Choose the equation(s) or situation: Use Newton’s second law to describe the forces acting on the bicycle. Fnet = m anet = −Fk Use the definition of frictional force to express Fk in terms of the coefficient of friction. Fk = mk Fn = mk (mg) Rearrange the equation(s) to isolate the unknown(s): m ane ...

Chapter 12 Problems

... spring of force constant k = 8.25 kN/m as shown in Figure P12.38. When no load is hung on the beam (Fg = 0), the length L is equal to 5.00 m. (a) Find the angle in this situation. (b) Now a load of Fg = 250 N is hung on the end of the beam. Temporarily ignore the extension of the spring and the ch ...

... spring of force constant k = 8.25 kN/m as shown in Figure P12.38. When no load is hung on the beam (Fg = 0), the length L is equal to 5.00 m. (a) Find the angle in this situation. (b) Now a load of Fg = 250 N is hung on the end of the beam. Temporarily ignore the extension of the spring and the ch ...

Mass Flow

... Free body diagrams • Free body diagram shows all external forces acting on the body. • Commonly used in statics, dynamics, and mechanics of materials • Steps to draw the free body diagram – Make a simplified drawing of the body in question – Draw all force vectors acting on it • Do not forget weigh ...

... Free body diagrams • Free body diagram shows all external forces acting on the body. • Commonly used in statics, dynamics, and mechanics of materials • Steps to draw the free body diagram – Make a simplified drawing of the body in question – Draw all force vectors acting on it • Do not forget weigh ...

2 equilibrium of force systems

... To define R completely, we locate its line of action by using the principle of moment. Choosing any point for center of moments, say point A, we get (clockwise positive). M A 150 301.8 504.2 757.2 54 210 540 804 kN.m Since, R r 804 r 804 170 4.73 m. Therefor ...

... To define R completely, we locate its line of action by using the principle of moment. Choosing any point for center of moments, say point A, we get (clockwise positive). M A 150 301.8 504.2 757.2 54 210 540 804 kN.m Since, R r 804 r 804 170 4.73 m. Therefor ...

Section 9.6: Work done on a many

... 9.3a, just one block is used to compress the spring while the other end of the spring is held against a wall. (a) Is the system comprising the block and the spring closed? (b) When the system is defined as being only the spring, is the work done by the block on the spring positive, negative, or zero ...

... 9.3a, just one block is used to compress the spring while the other end of the spring is held against a wall. (a) Is the system comprising the block and the spring closed? (b) When the system is defined as being only the spring, is the work done by the block on the spring positive, negative, or zero ...

Mastering Physics Answers

... Now draw a motion diagram, including all the elements listed in the problem-solving strategy. Use your diagram to answer the following questions. In interpreting the diagrams that follow, assume that the car is moving in a straight line to the right. Refer to this set of motion diagrams ...

... Now draw a motion diagram, including all the elements listed in the problem-solving strategy. Use your diagram to answer the following questions. In interpreting the diagrams that follow, assume that the car is moving in a straight line to the right. Refer to this set of motion diagrams ...

Classical Mechanics: a Critical Introduction

... (pushes and pulls caused by the action of one material object on another) and demons like “centrifugal force” (the tendency of an object moving in a circle to slip outwards) which must be expunged from the list of forces. An impatient reader may be annoyed by amount of space devoted to discussion of ...

... (pushes and pulls caused by the action of one material object on another) and demons like “centrifugal force” (the tendency of an object moving in a circle to slip outwards) which must be expunged from the list of forces. An impatient reader may be annoyed by amount of space devoted to discussion of ...

Everyday Forces and Laws of Motion

... • An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line) unless the object experiences a net external force. • In other words, when the net external force on an object is zero, the object’s acceleration (or t ...

... • An object at rest remains at rest, and an object in motion continues in motion with constant velocity (that is, constant speed in a straight line) unless the object experiences a net external force. • In other words, when the net external force on an object is zero, the object’s acceleration (or t ...

New St6 template Variatn6 - NSW Department of Education

... Ultimate tensile stress (UTS) is the maximum stress a material can withstand before it fails but not necessarily breaks. This is read from the top of the graphed line. UTS values are sometimes used in design work. Because the material has deformed plastically, it is necessary to compensate for this ...

... Ultimate tensile stress (UTS) is the maximum stress a material can withstand before it fails but not necessarily breaks. This is read from the top of the graphed line. UTS values are sometimes used in design work. Because the material has deformed plastically, it is necessary to compensate for this ...